Introduction. In this paper we consider some questions connected with the following problems from [5]:

VOL. LXX 1996 FASC. 2

ON THREE PROBLEMS FROM THE SCOTTISH BOOK CONNECTED WITH ORTHOGONAL SYSTEMS

BY

A. P L I C H K O

A. R A Z E N K O V (LVIV)

Introduction. In this paper we consider some questions connected with the following problems from [5]:

1. Problem of Mazur ([5, Problem 154]): Let (ϕ ⁿ ) be an orthogonal system consisting of continuous functions and closed in C.

(a) If f (t) ∼ a 1 ϕ 1 (t) + a 2 ϕ 2 (t) + . . . is the development of a given con- tinuous function f (t) and n 1 , n 2 , . . . denote the successive indices for which a n

6= 0, . . . , can one approximate f (t) uniformly by linear combinations of the functions ϕ n

(t), ϕ n

(t), . . .?

(b) Does there exist a linear summation method M such that the devel- opment of every continuous function f (t) in the system (ϕ n (t)) is uniformly summable by the method M to f (t)?

In [6] A. M. Olevski˘ı has given negative answers to both questions.

2. Problem of Banach ([5, Problem 86]): Given a sequence of functions (ϕ n (t)) which is orthogonal, normed, measurable, and uniformly bounded, can one always complete it, using functions with the same bound, to a sequence which is orthogonal, normed, and complete? Consider the case when infinitely many functions are necessary for completion.

This problem was first solved by S. Kaczmarz in [2]. Various solutions of this problem were found by B. S. Kashin, A. M. Olevski˘ı, S. V. Bochkarev and K. S. Kazarian [3, 4].

3. Problem of Mazur ([5, Problem 51]):

a) Is every set of functions, measurable in [0, 1] with the property that any two functions of the set are orthogonal, at most countable? (the func- tions are not assumed to be square-integrable!)

b) An analogous question for sequences: Is every set of sequences with the property that any two sequences (ε n ), (η n ) of this set are orthogonal, that is, P ∞

n=1 ε n η n = 0, at most countable?

1991 Mathematics Subject Classification: Primary 46B15.

[227]

It is stated in [5] that this problem was solved by Mazurkiewicz but there is no such remark in the xerox copy of the original manuscript we have.

Let X be a separable Banach space and let X ^∗ be its dual. A system x n , f n , x n ∈ X, f _n ∈ X ^∗ , n = 1, 2, . . . , ∞ is called biorthogonal if f m (x n ) = δ mn (Kronecker delta). A biorthogonal system is called fundamental (or complete) if its closed linear span [x n ] ^∞ _n=1 is equal to X, and total if for any non-zero element x ∈ X there is an index n such that f n (x) 6= 0. A fundamental and total biorthogonal system is called a Markushevich basis (an M -basis). A biorthogonal system is called a strong M -basis if x ∈ [f n (x)x n ] ^∞ _n=1 for every x in X. A system (x n ) is called a T -basis if there exists a regular summation method such that for every element x in X there exists a unique series P ∞

n=1 b n x n which is summable to x by this method.

We say that a Banach space X is densely embedded in a Banach space Y if X is a dense linear subspace of Y , it does not coincide with Y and there exists a positive constant C such that kxk Y ≤ Ckxk X for x ∈ X.

1. An answer to the first part of Mazur’s question [5, Problem 154]

follows from the following general proposition which is an improvement of results of Gurari˘ı and Johnson [1, 10].

Proposition 1. Let X be a separable Banach space which is densely embedded in a Hilbert space H. There exists a non-strong M -basis in X which is an orthogonal system in H.

For the proof we need three lemmas.

Lemma 1. Let X be a Banach space which is densely embedded in a Banach space Y and let E be a finite-codimensional closed subspace of Y . Then X ∩ E is densely embedded in E.

P r o o f. Let Z be a finite-dimensional complement to X ∩ E in X.

Then for every e ∈ E there exists a sequence x n + z n → e in Y -norm with x n ∈ X ∩ E and z _n ∈ Z. Since Z ∩ E = 0, Z and E are closed in Y and Z is finite-dimensional, we have x n → x and z _n → z as n → ∞, with x ∈ E, z ∈ Z and e = x + z. Thus z = 0 and x n → e as n → ∞, i.e. X ∩ E is densely embedded in E. If X ∩ E = E, then X = (X ∩ E) + Z = Y . Therefore X ∩ E 6= E.

Lemma 2. Let X be a Banach space which is densely embedded in a Hilbert space H. For any ε > 0 there exist x and x ⁰ in X such that kxk X = kx ⁰ k _X = 1, kx − x ⁰ k _X < ε and x⊥x ⁰ in H.

P r o o f. Let k k be the norm in X and k k H be the norm in H. Without

loss of generality we may suppose that there exists u in X such that kuk =

kuk _H = 1. Let E be the orthogonal complement of u in H. Then codim E

= 1. It follows from Lemma 1 that X ∩ E is dense in E and this embedding is not an isomorphism. Hence we may choose v in X ∩ E such that kvk H = 1 and a := kvk is sufficiently large. Put x = v + u, x ⁰ = v − u, x = x/kxk and x ⁰ = x ⁰ /kx ⁰ k. Then (x, x ⁰ ) = (v + u, v − u) = kvk H − kuk _H = 0, hence x⊥x ⁰ in H. It is easy to see that a − 1 < kxk, kx ⁰ k < a + 1. This implies that

| kxk − kx ⁰ k | ≤ 2 and kxk · kx ⁰ k ≥ (a − 1) ² . Then kx − x ⁰ k = k kx ⁰ kx − kxkx ⁰ k

kxk · kx ⁰ k ≤ k(kx ⁰ k − kxk)v + (kx ⁰ k + kxk)uk (a − 1) ²

≤ 2kvk + 2(a + 1)kuk

(a − 1) ² ≤ 2a + 2(a + 1) (a − 1) ² ,

i.e. choosing a sufficiently large we may obtain kx − x ⁰ k less than any pre- assigned ε.

Lemma 3. Let X be a Banach space which is densely embedded in a Hilbert space H. Let (ϕ n ) be a system which is fundamental in X and orthogonal in H. Then (ϕ n ) is an M -basis in X.

P r o o f. Since (ϕ n ) is orthogonal in H, there exist functionals (ϕ ^∗ _n ) ⊂ H ^∗ biorthogonal to (ϕ n ). Since X is densely embedded in H, H ^∗ is embedded in X ^∗ and dense in the weak ^∗ topology, hence (ϕ ^∗ _n ) is a total system on X, and therefore (ϕ n ) is an M-basis in X.

P r o o f o f P r o p o s i t i o n 1. Let (y n ) be some M-basis in X and let (ε n ) be a sequence of positive scalars such that lim n ε n = 0. We proceed by induction. In the first step we put z 1 = y 1 and choose x 1 and x ⁰ ₁ in X ∩ y ^⊥ ₁ which satisfy the conclusion of Lemma 2 with ε = ε 1 . In the nth step we put Y n−1 = (y i , x i , x ⁰ _i ) ⁿ⁻¹ _i=1 , take z n ∈ lin(Y n−1 , y n ) with z n ⊥Y n−1 and choose x n

and x ⁰ _n in X ∩ (Y n−1 ∪ {y _n }) ^⊥ which satisfy Lemma 2 with ε = ε n . Then the subspaces X 1 = [x n , z n ] ^∞ _n=1 and X 2 = [x ⁰ _n ] ^∞ _n=1 are quasi-complementary but not complementary in X and orthogonal in H. It is known (see [8] for example) that we can choose a subspace X ₁ ⁰ of X 1 such that dim X 1 /X ₁ ⁰ = 1 and so that X ₁ ⁰ and X 2 remain quasi-complementary in X. Take a system (u n ) which is complete in X ₁ ⁰ and orthogonalize it in H. We get a system (v n ) ⊂ X ₁ ⁰ for which all conditions of Lemma 3 are valid, hence (v n ) is an M- basis in X ₁ ⁰ , orthogonal in H. Put ϕ 2n−1 = v n and ϕ 2n = x n for n = 1, 2, . . . Then (ϕ n ) is an M-basis in X, it is orthogonal in H by Lemma 3, but it is not a strong M-basis because [ϕ 2n−1 ] ^∞ _n=1 ⊂ X ₁ ⁰ and ([ϕ 2n ] ^∞ _n=1 ) ^⊥ ⊃ X ₁ .

R e m a r k. Since every T-basis (summation basis) is a strong M-basis

(see [11, p. 357]), there exists an M-basis in X, orthogonal in H, which is

not a T-basis in X. In the case when X has a conditional basis which is

orthogonal in H, a negative answer to the second part of Mazur’s question

[5, Problem 154] can be obtained significantly simpler than in the article of

A. M. Olevski˘ı [6]. Such bases exist in L p , p > 2 (trigonometric system), and in C (Franklin system). We will show that such bases exist in some symmetric function spaces which are embedded in L 2 .

Proposition 2. Let X be a Banach space densely embedded in a Hilbert space H and suppose that X has a conditional basis orthogonal in H. Then there exists a strong M-basis in X, orthogonal in H, which is not a T-basis in X.

P r o o f. This easily follows from the fact that every conditional basis has a permutation which is not a T-basis (see [11, p. 357]). It is clear that the rearranged system remains an M-basis and orthogonal in H.

The following statement is well known (see [9, p. 31], for example).

Lemma 4. No orthonormal basis (x n (t)) ^∞ _n=1 in L 2 (0, 1) with |x n (t)| ≡ 1 for all n can be an unconditional basis of a symmetric space E on (0, 1) different from L 2 .

Let E be a symmetric function space, let p E and q E be its Boyd indices (see e.g. [9, p. 27] for definition). It is known that the Walsh system is a basis in L p , 1 < p < ∞. If 1 ≤ p E ≤ q E < ∞, then E is an interpolation space between L p

and L q

([9, p. 27]). The above observations imply that the Walsh system is a conditional basis in E when 2 < p E ≤ q _E < ∞.

2. The following proposition gives, in particular, a negative answer to Banach’s question [5, Problem 86].

Proposition 3. Let X be a Banach space which is densely embedded in a Hilbert space H and this embedding is not compact. Then there exists a sequence (ϕ n ) such that

(i) (ϕ n ) is bounded in X;

(ii) (ϕ n ) is orthogonal in H;

(iii) (ϕ n ) admits no extension to a fundamental and orthogonal sequence in H, using elements from X;

(iv) the closed linear span of (ϕ n ) in H has an infinite codimension in H.

We need two lemmas for the proof.

Lemma 5. Let X be a Banach space which is densely embedded in a Hilbert space H and the embedding is not compact. Then there exists a positive scalar a such that for any finite-codimensional closed subspace E ⊂ H there exists x ∈ X ∩ E ^⊥ such that kxk X ≤ akxk _X .

P r o o f. Suppose the converse. Then for every a there exists a finite- codimensional subspace E ⊂ H with kxk X ≥ akxk _H for every x ∈ X ∩ E.

We will show that this implies the compactness of the embedding of X in H.

We need to show that for every ε > 0 there exists a finite cover of B(X) (the unit ball of X) by balls S 1 , . . . , S m in H with radius ε. It follows from the assumption that B(X) ∩ E ⊆ εB(H) if ε = 1/a. Compactness of the embedding now easily follows from the fact that X ∩ E is closed and finite-codimensional.

Lemma 6. Let X be a Banach space which is densely embedded in a Hilbert space H. Let E be a finite-codimensional closed subspace of H, let ε > 0 and v ∈ X. Then there exists y ∈ X ∩E such that d(v, lin(E ^⊥ , y)) < ε, where d means the distance in H.

P r o o f. Decompose v in H as v = v ^∗ + v ^∗∗ , where v ^∗ ∈ E and v ^∗∗ ∈ E ^⊥ . Hence

d(v, lin(E ^⊥ , y)) = inf{kv − zk H : z ∈ lin(E ^⊥ , y)}

= inf{(kv ^∗ − λyk ² + kv ^∗∗ − uk ² ) ^1/2 : λ ∈ R, u ∈ E ^⊥ }

= inf{kv ^∗ − λyk H : λ ∈ R}.

Since X ∩ E is densely embedded in E by Lemma 1, v ^∗ can be approximated arbitrarily closely by an element y from X ∩ E.

P r o o f o f P r o p o s i t i o n 3. The proof is a modification of arguments from [7]. The reasoning uses the orthogonal transformation of A. M. Olevski˘ı and takes into account results from [8].

Let (v n ) ^∞ _n=1 ⊂ X be a complete sequence in H such that each element is repeated infinitely many times. Let (ε n ) ^∞ _n=1 be a sequence of positive scalars such that lim n ε n = 0. By [8] there exists a closed infinite-dimensional subspace Z in H such that Z ∩ X = 0. We proceed by induction. Let a be the constant from Lemma 5. For elements (z i , x i , y i ) ⁿ _i=1 ⊂ H we put H n = lin(z i , x i , y i ) ⁿ _i=1 . In the first step we use Lemma 5 to find z 1 ∈ Z, z 1 6= 0, and x ₁ ∈ X with x ₁ ⊥z ₁ , kx 1 k _H = 1 and kx 1 k _X ≤ a. Next we use Lemma 6 to choose y 1 ∈ X such that y 1 ∈ (z 1 , x 1 ) ^⊥ , ky 1 k H = 1 and d(v 1 , H 1 ) < ε 1 . In the nth step we take z n ∈ Z ∩ H _n−1 ^⊥ , z n 6= 0, choose x n ∈ X ∩ H _n−1 ^⊥ ∩ z ^⊥ _n such that kx n k _H = 1 and kx n k _X ≤ a and choose y n ∈ X ∩ H _n−1 ^⊥ ∩ (z n , x n ) ^⊥ such that ky 1 k H = 1 and d(v n , H n ) < ε n .

Now we rearrange the sequence (y n ) and relabel it as (ψ i

) ^∞ _m=1 , where

(i m ) ^∞ _m=1 is an increasing sequence such that for every m, i m − i _m−1 = 2 ^s

,

where the positive integer s m is chosen to satisfy 2 ^−s

^/2 kψ i

k X < 2 ^−m .

We relabel (x n ) ^∞ _n=1 using the remaining positive integers to get the sequence

(ψ i : i 6∈ (i m ) ^∞ _m=1 ). Let us apply for each block (ψ i : i m−1 < i ≤ i m ) the

orthogonal transformation of Olevski˘ı [7]. We obtain a sequence (ϕ k ) ^∞ _k=1

which is bounded in X and orthonormal in H. The closed linear span of

(ϕ k ) ^∞ _k=1 in H coincides with the closed linear span of (x n , y n ) ^∞ _n=1 in H. It is

clear that the subspace [z n ] ^∞ _n=1 is an orthogonal complement to this closed

linear span.

3. In this section we will answer Mazur’s question [5, Problem 51]. First we consider the discrete variant. We need the following known lemma ([11, p. 208]).

Lemma 7. Let N be a countable set. Then there exists a family {M α } _α∈A of subsets of N with the following properties:

(i) The index set A has cardinality continuum.

(ii) Each set M α is infinite.

(iii) M α ∩ M _β is finite for α 6= β.

P r o o f. Let N be the set of all rational numbers in (0, 1), A be the set of all irrational numbers in (0, 1) and, for each α ∈ A, let M α be an arbitrary infinite sequence in N converging to α.

Proposition 4. There exist continuum many sequences x ^α = (x ^α ₁ , x ^α ₂ , . . .), α ∈ A, such that x ^α _n = 0, 1, or −1 for every n and α, and for α 6= β the series P ∞

n=1 x ^α _n x ^β _n contains a finite number of non-zero terms and its sum is equal to zero.

P r o o f. In a countable set N 0 choose continuum many non-empty sub- sets M α , α ∈ A, such that M α ∩ M β is a finite set for α 6= β, by Lemma 7.

Put x ^α = (x ^α ₁ , x ^α ₂ , . . .), where x ^α _n = 1 if n ∈ M α , and x ^α _n = 0 if n 6∈ M α . We have constructed continuum many sequences x ^α so that for every α 6= β the series P ∞

n=1 x ^α _n x ^β _n is a finite sum.

Now we represent A as a dyadic tree A = A 1 t A ₂ , A 1 = A 3 t A ₄ , A 2 = A 5 t A 6 , . . . , where t denotes disjoint union, by the scheme:

A

A 1 A 2

A 3 A 4 A 5 A 6

A 7 A 8 A 9 A 10 A 11 A 12 A 13 A 14

 

 



== == =

ppp ppp ppp pp

~~ ~~ ~ @@@ @@

OOOOO OOOOO OO

 



.. .. . 

 

// // / 

 

// // / 

 

// // /

We make this representation in such a way that

(∗) every chain (A k

) ^∞ _i=1 has one-point intersection T ∞ i=1 A k

.

We shall add to N 0 a countable number of countable sets N i , i = 1, 2, . . . , and shall complete the definition of our sequences on S ∞

i=1 N i by 0, 1, and

−1 so that in the ith step for α 6= β the series S(α, β, i) := X

n∈ ∪

^N

x ^α _n x ^β _n

will contain a finite number of non-zero terms; if S(α, β, i) = 0 then S(α, β, j) = 0 for j > i; and for every α 6= β there exists i such that S(α, β, i) = 0.

First step. Let N 1 be a copy of N 0 and ϕ 1 : N 0 → N ₁ be an identifying map. Put

x ^α _ϕ

_(n) =  x ^α _n if α ∈ A 1 ,

−x ^α _n if α ∈ A 2 , n ∈ N 0 .

Then for every α, β the series S(α, β, 1) has a finite number of non-zero terms and its sum is zero for α ∈ A 1 , β ∈ A 2 .

Second step. Let N 2 be a copy of N 0 ∪ N ₁ and ϕ 2 : N 0 ∪ N ₁ → N ₂ be an identifying map. Put

x ^α _ϕ

_(n) =

( x ^α _n if α ∈ A 3 ,

−x ^α _n if α ∈ A 4 , 0 if α 6∈ A 1 ,

n ∈ N 0 ∪ N ₁ .

Then for every α, β the series S(α, β, 2) has a finite number of non-zero terms, its sum is zero for α ∈ A 3 , β ∈ A 4 , and also for α ∈ A 1 , β ∈ A 2 , since it is then equal to S(α, β, 1).

Third step. Let N 3 be a copy of N 0 ∪N ₁ ∪N ₂ and ϕ 3 : N 0 ∪N ₁ ∪N ₂ → N ₃ be an identifying map. Put

x ^α _ϕ

_(n) =

( x ^α _n if α ∈ A 5 ,

−x ^α _n if α ∈ A 6 , n ∈ N 0 ∪ N ₁ ∪ N ₂ . 0 if α 6∈ A 2 ,

Then for every α, β the series S(α, β, 3) has a finite number of non-zero terms, its sum is zero for α ∈ A 5 , β ∈ A 6 , for α ∈ A 1 , β ∈ A 2 (being equal to S(α, β, 1)) and for α ∈ A 3 , β ∈ A 4 (being equal to S(α, β, 2)).

We have constructed our sequences so that in the ith step for α 6= β the series S(α, β, i) has a finite number of non-zero terms and if S(α, β, i) = 0 then S(α, β, j) = S(α, β, i) = 0 for j > i. Condition (∗) ensures that for any distinct α, β there exists i such that S(α, β, i) = 0.

An uncountable orthogonal system on an interval can be obtained as a result of the following transformation. We decompose (0, 1) into a countable union of disjoint sets (∆ n ) ^∞ _n=1 of positive measure, and for every sequence x = (x n ) we define a function f x (t) = x n /pµ(∆ n ) for t ∈ ∆ n . It is easy to see that if x ^α , α ∈ A, are the sequences from Proposition 4, then the set of functions f x

, α ∈ A, has the property desired in [5, Problem 51].

Acknowledgements. The authors are grateful to Prof. V. Gaposhkin

and Prof. A. Pe lczy´ nski for useful advice.

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Institute of Applied Problems of Mechanics and Mathematics Ukrainian Academy of Sciences

Naukova 3b

290053 Lviv, Ukraine Current address:

Department of Mathematics Pedagogical Institute Shevchenko St. 1

316050 Kirovograd, Ukraine

Re¸ cu par la R´ edaction le 5.5.1994;

en version modifi´ ee le 11.7.1995